Measure and draw angles accurately, construct triangles and other figures using a ruler and compasses, and understand how nets fold into three-dimensional shapes.
Subtopic 1
A protractor measures an angle in degrees. Most semicircular protractors show angles from \(0^\circ\) to \(180^\circ\), while a circular angle measurer may show a full \(360^\circ\). The vertex is where the two arms meet. In \(\angle PQR\), the middle letter Q is the vertex.
Acute: \(0^\circ<\theta<90^\circ\). Right: \(90^\circ\). Obtuse: \(90^\circ<\theta<180^\circ\). Straight: \(180^\circ\). Reflex: \(180^\circ<\theta<360^\circ\). Full turn: \(360^\circ\).
A protractor appears to show readings of \(62^\circ\) and \(118^\circ\). The angle is clearly obtuse. Which reading is correct?
Step 1: An obtuse angle is larger than \(90^\circ\).
Step 2: \(62^\circ\) is acute and is not sensible.
Answer: \(\boxed{118^\circ}\).
In \(\angle ABC\), which letter is the vertex? How would you position your protractor?
Step 1: The middle letter is the vertex, so it is B.
Step 2: Place the protractor centre at B and align the zero line with BA or BC.
An angle is a little less than a right angle. Is a protractor reading of \(84^\circ\) or \(96^\circ\) more plausible?
Step 1: A right angle is \(90^\circ\).
Step 2: A little less than \(90^\circ\) means acute.
Answer: \(\boxed{84^\circ}\).
The smaller angle between two rays is \(52^\circ\). Find the reflex angle.
Step 1: Angles around a full turn total \(360^\circ\).
Step 2: \(360^\circ-52^\circ=\boxed{308^\circ}\).
Draw a ray OA of length 6 cm and construct \(\angle AOB=70^\circ\).
Step 1: Use a ruler to draw OA exactly 6 cm.
Step 2: Centre the protractor at O and place the zero line along OA.
Step 3: Make a small mark at \(70^\circ\), remove the protractor, and rule OB through O and that mark.
Check: Measure the completed angle; it should be about \(70^\circ\).
Draw \(\angle XYZ=135^\circ\) using a protractor.
Step 1: Draw ray YX and place the protractor centre at Y.
Step 2: Read the scale that starts at zero on YX; mark \(135^\circ\).
Step 3: Draw YZ through the mark. Check that the angle looks obtuse, not acute.
A diagram shows a reflex angle at O. The smaller unshaded angle is measured as \(47^\circ\). A student writes \(47^\circ\) for the shaded reflex angle. Explain the error and calculate the correct angle.
Step 1: The student measured the smaller, unshaded angle instead of the angle requested.
Step 2: Reflex angle \(=360^\circ-47^\circ\).
Answer: \(\boxed{313^\circ}\), which is larger than \(180^\circ\), as expected.
A student draws a nominal \(120^\circ\) angle, but the protractor scale aligned with the correct starting ray reads \(119^\circ\). What should be recorded from the actual drawing?
Step 1: The requested target and the measured result are different.
Step 2: Record the measurement of the completed drawing as approximately \(\boxed{119^\circ}\), to the nearest degree.
Exam habit: If asked to construct \(120^\circ\), improve the drawing until it matches the target.
Subtopic 2
When all three side lengths of a triangle are known, its third vertex can be located using two compass arcs. This is called a side–side–side (SSS) construction. It requires a ruler and compasses; you do not need a protractor to find the third vertex. Draw arcs lightly and leave construction marks visible.
Construct triangle ABC where \(AB=5\text{ cm}\), \(AC=3\text{ cm}\) and \(BC=4\text{ cm}\).
Step 1: Draw AB as a 5 cm base.
Step 2: With centre A and radius 3 cm, draw an arc.
Step 3: With centre B and radius 4 cm, draw an arc meeting the first at C.
Step 4: Rule AC and BC. Check lengths and leave arcs visible.
You draw XY = 7 cm. You are given XZ = 5 cm and YZ = 6 cm. State the compass settings and centres.
Step 1: Draw XY as 7 cm.
Step 2: Set compasses to 5 cm from X.
Step 3: Set compasses to 6 cm from Y.
Their intersection is Z.
Construct an equilateral triangle PQR with side length 4.8 cm.
Step 1: Draw PQ = 4.8 cm.
Step 2: Use a compass opening of 4.8 cm and draw arcs centred on P and Q.
Step 3: Mark intersection R; join PR and QR.
Check: All three sides measure 4.8 cm.
Construct triangle ABC with \(AB=6\text{ cm}\), \(AC=5\text{ cm}\), \(BC=5\text{ cm}\).
Step 1: Draw AB = 6 cm.
Step 2: Draw a 5 cm arc from A and a 5 cm arc from B.
Step 3: Mark their intersection C and join the sides.
Equal sides AC and BC confirm the triangle is isosceles.
Can you construct a triangle whose sides are 2 cm, 3 cm and 6 cm? Explain.
Step 1: The maximum reach of the two shorter sides together is \(2+3=5\text{ cm}\).
Step 2: That is less than the required 6 cm side, so the arcs will not meet.
Answer: No triangle is possible. This is a useful construction check.
A base AB is 8 cm, with \(AC=6\text{ cm}\) and \(BC=5\text{ cm}\). Why can the third vertex appear on either side of AB?
Step 1: The two arcs may intersect once above AB and once below it.
Step 2: Both intersections give triangles with the same three side lengths.
They are mirror images. Use the side specified by the question if it asks for one orientation.
Construct rhombus ABCD with diagonal \(AC=8\text{ cm}\) and every side 5 cm. Explain how to find B and D.
Step 1: Draw AC = 8 cm.
Step 2: Draw 5 cm arcs centred at A and C.
Step 3: Mark the two arc intersections B and D, on opposite sides of AC.
Step 4: Join A–B–C–D–A. Each side is 5 cm by the compass construction.
A rhombus has four sides of 5 cm and diagonal AC = 8 cm. What length should its other diagonal BD measure?
Step 1: Make the rhombus using 5 cm compass arcs from the ends of AC = 8 cm.
Step 2: Measure BD carefully from the finished drawing.
Check: BD should measure approximately \(\boxed{6\text{ cm}}\) to the nearest millimetre. This is a practical measurement check, not an additional construction step.
Construct triangle DEF with \(DE=7.2\text{ cm}\), \(DF=5.4\text{ cm}\), \(EF=6.1\text{ cm}\). Which measurements should be taken at the end?
Step 1: Draw DE = 7.2 cm.
Step 2: Draw an arc from D with radius 5.4 cm.
Step 3: Draw an arc from E with radius 6.1 cm, meeting the first at F.
Step 4: Rule DF and EF. Verify DE = 7.2 cm, DF = 5.4 cm and EF = 6.1 cm, ideally to 1 mm; keep arcs visible.
Subtopic 3
A net is a flat arrangement of faces that folds to form a three-dimensional solid without overlapping or leaving gaps. A net must include the right number and shape of faces, and those faces must be attached in a workable arrangement.
How many square faces must a closed cube net contain?
Step 1: A cube has six faces.
Step 2: Each is a square.
Answer: \(\boxed{6}\) squares, attached in a net that folds correctly.
A triangular-based pyramid has a triangular base and three triangular side faces. How many triangular faces are in its net?
Step 1: There is 1 base triangle.
Step 2: Add 3 side triangles.
Answer: \(1+3=\boxed{4}\) triangular faces.
In the labelled cuboid net shown above, which labelled corners join point A after folding?
Step 1: Imagine wrapping the four side rectangles around the cuboid.
Step 2: Fold the top and bottom faces into position.
Answer: The source diagram shows corners A, C and G meeting at one vertex. Thus C and G join A.
In the six arrangements labelled a–f above, which three fold to make a cuboid?
Step 1: Check that all six faces can fold without overlapping.
Step 2: Test how narrow side faces attach around the main rectangles.
Answer from the source diagram: a, b and e fold correctly; c, d and f do not.
For length \(l\), width \(w\), height \(h\):
\[\text{Total surface area}=2(lw+lh+wh)\]
\[\text{Volume}=lwh\]
Surface area is measured in square units (e.g. cm²); volume is measured in cubic units (e.g. cm³). In an open box, omit any missing face from surface area.
A closed cuboid measures 4 cm by 3 cm by 2 cm. Find its total surface area.
Step 1: Face-pair areas are \(4\times3=12\), \(4\times2=8\), and \(3\times2=6\) cm².
Step 2: There are two of each face.
Answer: \(2(12+8+6)=\boxed{52\text{ cm}^2}\).
A net folds into a cuboid measuring 5 cm by 4 cm by 3 cm. Find the volume.
Step 1: Use \(V=lwh\).
Step 2: \(5\times4\times3=\boxed{60\text{ cm}^3}\).
A triangular prism has a triangular cross-section with sides 3 cm, 4 cm and 5 cm, and a length of 8 cm. Describe its net.
Step 1: The prism has 2 congruent triangular ends, each 3–4–5.
Step 2: Its 3 lateral faces are rectangles sized \(3\times8\), \(4\times8\) and \(5\times8\) cm.
Answer: Draw a strip of three correctly matched rectangles and attach a triangle to each open end.
A right triangular prism has end sides 3 cm, 4 cm and 5 cm and length 10 cm. Find its total surface area.
Step 1: The two right-triangle ends total \(2\times\frac12\times3\times4=12\text{ cm}^2\).
Step 2: Lateral area \(=(3+4+5)\times10=120\text{ cm}^2\).
Answer: \(120+12=\boxed{132\text{ cm}^2}\).
A box with no lid measures 12 cm long, 8 cm wide and 5 cm high. Find the area of card needed (ignore joins).
Step 1: Base area \(12\times8=96\text{ cm}^2\).
Step 2: Two long sides: \(2(12\times5)=120\text{ cm}^2\).
Step 3: Two short sides: \(2(8\times5)=80\text{ cm}^2\).
Answer: \(96+120+80=\boxed{296\text{ cm}^2}\). Do not include the missing lid.
The same 12 cm by 8 cm by 5 cm box is drawn as a net at a scale of 1 cm on paper to 4 cm in real life. Give the dimensions of the base and side flaps on the drawing.
Step 1: Divide every actual length by 4.
Step 2: Base is \(12\div4=3\text{ cm}\) by \(8\div4=2\text{ cm}\).
Step 3: Height of each side flap \(5\div4=1.25\text{ cm}\).
Answer: A \(3\times2\) cm base, with two \(3\times1.25\) cm flaps and two \(2\times1.25\) cm flaps.
Work through the easy questions first, then medium, then exam-style questions. The later questions are original Cambridge-style practice, not reproduced past-paper questions. Answers are hidden until you open Show Answer.
Classify \(35^\circ\) as acute, right, obtuse or reflex.
Step 1: \(35^\circ<90^\circ\).
Answer: Acute.
Classify \(146^\circ\).
Step 1: \(90^\circ<146^\circ<180^\circ\).
Answer: Obtuse.
In \(\angle MNO\), name the vertex.
Step 1: The middle letter names the vertex.
Answer: N.
A protractor shows \(56^\circ\) on one scale and \(124^\circ\) on the other. The drawn angle is obtuse. What should you record?
Step 1: An obtuse angle exceeds \(90^\circ\).
Answer: \(\boxed{124^\circ}\).
The smaller angle at a point is \(41^\circ\). Calculate its reflex angle.
Step 1: Reflex angle \(=360^\circ-\text{smaller angle}\).
Step 2: \(360-41=\boxed{319^\circ}\).
A reflex angle is \(296^\circ\). Find the smaller angle.
Step 1: Smaller angle \(=360^\circ-296^\circ\).
Answer: \(\boxed{64^\circ}\).
Draw a line AB of length 7.5 cm with a ruler. State the length in millimetres.
Step 1: \(1\text{ cm}=10\text{ mm}\).
Answer: \(7.5\times10=\boxed{75\text{ mm}}\). Use the 75 mm mark when drawing.
Draw \(\angle PQR=65^\circ\) using a protractor. Explain the steps.
Step 1: Draw ray QP.
Step 2: Place the protractor centre on Q and its zero line on QP.
Step 3: Mark 65° on the correct scale, draw QR through the mark.
Check: Remeasure to the nearest 1°.
Draw an angle of \(142^\circ\). State how you know you have read the correct protractor scale.
Step 1: Draw a base ray and centre the protractor at the vertex.
Step 2: Read from 0 on the ray to 142°.
Check: The angle is obtuse; a reading of 38° would indicate the wrong scale.
An angle is described as reflex. A learner measures the smaller opening as \(73^\circ\) and writes that as the answer. Correct the work and explain the error.
Step 1: A reflex angle is the larger turn, not the smaller angle.
Step 2: \(360^\circ-73^\circ=\boxed{287^\circ}\).
Error: The learner recorded the acute opening instead of subtracting from a full turn.
Two arms of an angle overlap on the start ray at a full turn. What is the angle of one complete rotation, and what will a semicircular protractor display if positioned on the final ray?
Step 1: A complete rotation is \(360^\circ\).
Step 2: A protractor aligned with two overlapping rays would show a zero opening, not the number of rotations.
Answer: The turn is \(\boxed{360^\circ}\), while the final opening is \(0^\circ\).
Draw a ray OA 6 cm long, construct an obtuse angle \(\angle AOB=118^\circ\), and explain two checks you would make.
Step 1: Draw OA = 6 cm.
Step 2: Centre the protractor at O and mark 118° from ray OA. Draw OB.
Checks: OA measures 60 mm; the constructed opening measures about 118°, which looks obtuse.
A drawn angle has a smaller opening of \(154^\circ\). Find the reflex opening and explain why the answer is sensible.
Step 1: Subtract from a full turn: \(360-154=206\).
Answer: \(\boxed{206^\circ}\).
Check: 206° is more than 180° and less than 360°, as a reflex angle must be.
For triangle ABC, AB = 6 cm, AC = 4 cm and BC = 5 cm. If AB is the base, state the two compass openings and their centres.
Step 1: Draw AB = 6 cm.
Step 2: From A draw radius 4 cm; from B draw radius 5 cm.
The intersection is C.
Why should construction arcs be left visible?
Step 1: They show how distances were measured and the vertex located.
Answer: The examiner can check the construction method, not merely the final shape.
Construct an equilateral triangle with sides 5 cm.
Step 1: Draw the 5 cm base.
Step 2: Draw 5 cm arcs from both endpoints.
Step 3: Join the arc intersection to each endpoint. Each side must measure 5 cm.
Construct triangle PQR with PQ = 7 cm, PR = 4 cm and QR = 6 cm.
Step 1: Draw PQ = 7 cm.
Step 2: From P use a 4 cm arc; from Q use a 6 cm arc.
Step 3: Mark R at their intersection and rule PR, QR. Check all lengths.
Construct an isosceles triangle with base 8 cm and the other two sides 6 cm.
Step 1: Draw an 8 cm base AB.
Step 2: Use compasses set to 6 cm at both A and B.
Step 3: Join the meeting point C to each end; AC = BC = 6 cm.
A pupil sets the compasses to 3 cm from A and 4 cm from B for a base AB of 9 cm. Explain why no triangle is formed.
Step 1: The arcs must meet for the triangle to exist.
Step 2: \(3+4=7<9\).
Answer: The arcs cannot reach each other; no triangle is possible.
Triangle ABC is constructed with sides 5 cm, 6 cm and 7 cm. If its arcs intersect both above and below the base, are both positions valid?
Step 1: Both intersections are the specified distances from the base endpoints.
Step 2: Therefore both positions form valid 5–6–7 triangles.
They are reflections of one another across the base.
Explain why a ruler alone is not the correct method for locating the third vertex of a triangle from three lengths.
Step 1: A ruler measures distance from one endpoint but does not give all possible points at that distance easily.
Step 2: Each compass arc represents points an exact distance from its centre.
The intersection meets both side-length conditions simultaneously.
Construct triangle XYZ with XY = 8.2 cm, XZ = 5.7 cm and YZ = 6.9 cm. Describe how to verify your result.
Step 1: Draw XY = 8.2 cm.
Step 2: Draw the 5.7 cm arc from X and 6.9 cm arc from Y.
Step 3: Join their intersection to X and Y.
Check: Remeasure XY, XZ and YZ to about 1 mm and leave arcs.
A rhombus has sides 5 cm and diagonal AC = 6 cm. Explain how to construct its four vertices using a ruler and compasses.
Step 1: Rule AC = 6 cm.
Step 2: Draw arcs of radius 5 cm centred at A and C.
Step 3: Mark the intersections B and D on opposite sides of AC.
Step 4: Rule AB, BC, CD and DA; verify each side is 5 cm.
In the rhombus in the previous question, find the length of diagonal BD as a check on the drawing.
Step 1: Construct the rhombus by drawing 5 cm arcs from the endpoints of AC = 6 cm.
Step 2: Measure BD on the paper drawing using a ruler.
Check: BD should be close to \(\boxed{8\text{ cm}}\). Keep the construction arcs visible.
An isosceles triangle uses the two distinct side lengths 5 cm and 6 cm, with one length occurring twice. Explain the two different triangles that can be constructed.
Step 1: If the repeated sides are 6 cm, use base 5 cm and two 6 cm arcs.
Step 2: If the repeated sides are 5 cm, the third side must be 6 cm and use two 5 cm arcs.
Answer: These are two distinct types of isosceles triangle possible using lengths 5 cm and 6 cm (not just mirror placements).
Why is a triangle with lengths 4 cm, 7 cm and 11 cm not a genuine triangle with three distinct vertices?
Step 1: The two shorter lengths add to \(4+7=11\) cm.
Step 2: They reach the third endpoint only along the same straight line.
Answer: The would-be vertices are collinear; the area is zero, so it is not a proper triangle.
Draw a 5 cm by 5 cm rhombus whose diagonal AC is 8 cm. State which instruments locate B and D, and which joins them.
Step 1: Use a ruler to draw AC = 8 cm.
Step 2: With compasses set at 5 cm, draw arcs centred at A and C on both sides, giving B and D.
Step 3: Use a ruler to join consecutive vertices. Leave arcs visible.
How many faces, edges and vertices does a cuboid have?
Step 1: Identify its six rectangular faces.
Answer: \(\boxed{6}\) faces, \(\boxed{12}\) edges and \(\boxed{8}\) vertices.
What shapes form the faces of a cube net?
Step 1: Every face of a cube is a square.
Answer: Six equal squares, arranged to fold without overlap.
How many triangles form the net of a triangular-based pyramid?
Step 1: Count 1 base triangle and 3 sloping faces.
Answer: \(1+3=\boxed{4}\) triangles.
A closed cuboid has dimensions 6 cm by 4 cm by 2 cm. Name the three different face sizes.
Step 1: Pair the three dimensions.
Answer: \(6\times4\), \(6\times2\) and \(4\times2\) cm rectangles, two of each.
A closed cuboid has dimensions 3 cm by 2 cm by 1 cm. Find its total surface area.
Step 1: Areas of different face pairs: 6, 3 and 2 cm².
Step 2: \(2(6+3+2)=\boxed{22\text{ cm}^2}\).
A cuboid is 7 cm long, 5 cm wide and 3 cm high. Find its volume.
Step 1: \(V=lwh\).
Step 2: \(7\times5\times3=\boxed{105\text{ cm}^3}\).
Explain why six rectangles with the correct dimensions may still fail to form a cuboid net.
Step 1: Number and sizes of faces are necessary, but not sufficient.
Step 2: Faces might overlap, meet at incompatible edges or fail to close.
Check: Mentally fold or make a paper model.
A triangular prism has a triangular cross-section with side lengths 5 cm, 5 cm and 6 cm, and prism length 9 cm. List the rectangles in its net.
Step 1: Each triangle side pairs with the 9 cm prism length.
Answer: Rectangles \(5\times9\), \(5\times9\) and \(6\times9\) cm, plus two equal triangles.
Draw a net for a cuboid 4 cm by 3 cm by 2 cm. How many rectangles of each dimension should appear?
Step 1: A cuboid has three pairs of opposite faces.
Answer: Two 4×3 rectangles, two 4×2 rectangles and two 3×2 rectangles.
Check: Arrange and connect them so the net folds without overlap.
The source textbook’s labelled cuboid net shows A, C and G meeting at one corner. Which labels meet at A?
Step 1: Fold the side faces, then the top.
Answer: \(\boxed{C\text{ and }G}\) meet A.
A box with no lid is 15 cm long, 10 cm wide and 4 cm high. Calculate the area of card in its net, excluding tabs.
Step 1: Base \(15\times10=150\text{ cm}^2\).
Step 2: Long sides \(2\times15\times4=120\text{ cm}^2\).
Step 3: Short sides \(2\times10\times4=80\text{ cm}^2\).
Answer: \(150+120+80=\boxed{350\text{ cm}^2}\).
A closed cuboid is 8 cm by 5 cm by 4 cm. Calculate both its total surface area and volume.
Step 1: \(SA=2(8\times5+8\times4+5\times4)=2(40+32+20)=\boxed{184\text{ cm}^2}\).
Step 2: \(V=8\times5\times4=\boxed{160\text{ cm}^3}\).
Check: Surface area has square units; volume has cubic units.
A triangular prism has right-triangle ends with legs 6 cm and 8 cm and hypotenuse 10 cm. Its length is 12 cm. Find its total surface area.
Step 1: Two triangular ends: \(2\times\frac12\times6\times8=48\text{ cm}^2\).
Step 2: Three rectangles: \((6+8+10)\times12=288\text{ cm}^2\).
Answer: \(48+288=\boxed{336\text{ cm}^2}\).
The prism in the previous question has a uniform right-triangle cross-section. Calculate its volume.
Step 1: Area of cross-section \(=\frac12\times6\times8=24\text{ cm}^2\).
Step 2: Multiply by its length 12 cm.
Answer: \(V=24\times12=\boxed{288\text{ cm}^3}\).
A square-based pyramid has a square base and four triangular side faces. Describe a valid net and state the number of faces.
Step 1: Draw one square base.
Step 2: Join one correctly sized triangle to each of the four square sides.
Answer: \(1+4=\boxed{5}\) faces in a workable net.
A cube has edge length 4.5 cm. A net is cut from card. Calculate the total area of card excluding tabs.
Step 1: Each square face area \(=4.5^2=20.25\text{ cm}^2\).
Step 2: Six faces give \(6\times20.25=\boxed{121.5\text{ cm}^2}\).
A cuboid is 12 cm by 8 cm by 6 cm. A lid is removed. Calculate the surface area of the remaining five faces and the volume of the original cuboid.
Step 1: Base \(12\times8=96\).
Step 2: Sides \(2(12\times6)+2(8\times6)=144+96=240\).
Five-face area: \(96+240=\boxed{336\text{ cm}^2}\).
Volume: \(12\times8\times6=\boxed{576\text{ cm}^3}\).
Source note: Lesson headings and mathematical topic coverage follow Chapter 33 of the supplied Hodder Cambridge Mathematics textbook (printed pp. 287–294). Figures are crops from that chapter. The explanations, worked examples and exam-style questions here are newly written IGCSE360 teaching material, not official past-paper questions.